Question

The vapor pressure of a substance describes how readily molecules at the surface of the substance enter the gaseous phase. At the boiling point of a liquid, the liquid's vapor pressure is equal to or greater than the atmospheric pressure exerted on the surface of the liquid. Since the atmospheric pressure at higher elevations is lower than at sea level, the boiling point of water decreases as the elevation increases. The atmospheric pressure at sea level is 760 mmHg. This pressure decreases by 19.8 mmHg for every 1000-ft increase in elevation.

Elevation	Pressure
0 ft	760 mmHg
1000 ft	740.2 mmHg
2000 ft	720.4 mmHg

The boiling point of water decreases 0.05∘C for every 1 mmHg drop in atmospheric pressure.

A) What is the boiling point of water at an elevation of 6.00×10^3 ft ?

Answer 1

I think this is going to be straightforward to explain. Since the pressure drops 19.8 mm Hg for every 1000 ft increase, then for 6000 feet it decreases (6000/1000) x 19.8 = 118.8 mm Hg.

Since the boiling point decreases 0.05C for every 1 mm Hg drop, the boiling point will decrease 118.8 x 0.05 = 5.94 degrees C.
Therefore, the boiling point is 100C - 5.94C = 96.06 degrees C(your sig figs are actually should round off to 95 degrees C since the data point of 0.05 is only one sig fig; you might report both, and explain why you are doing so. You might help your teacher change the question for next time!).